Problem: Consider the parametric curve: $\begin{aligned} x&=t^{2} \\\\ y&=-4\tan(t) \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=-\dfrac{\pi}{6}$ to $t=\dfrac{\pi}{3}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{4t^2+16\sec^4(t)}\,dt$ (Choice B) B $\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{t^{2}-4\tan(t)}\,dt$ (Choice C) C $\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{t^4+16\tan^2(t)}\,dt$ (Choice D) D $\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{2t-4\sec^2(t)}\,dt$
Solution: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[t^{2}\right] \\\\ &=2t \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[-4\tan(t)\right] \\\\ &=-4\sec^2(t) \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{\left(2t\right)^2+\left(-4\sec^2(t)\right)^2}\,dt \\\\ &=\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{4t^2+16\sec^4(t)}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=-\dfrac{\pi}{6}$ to $t=\dfrac{\pi}{3}$ : $\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} \sqrt{4t^2+16\sec^4(t)}\,dt$